Sorry for the rookie question here but I was curious.
I was watching my CabFoward do its thing today and I actually had the wheels slip a bit on grade after haveing to stop it. I know how my model powered, but what about the real thing… Let me clarify…
The wheels are pushed by the rod and then pulled or are thy just pushed with main power and enough power to return the crank and rod to be pushed again? So the power only really hits the wheel say from 12 o’clock to 6 o’clock?
Or is there a pulling power to so the power is constant through the rotation?
Or, are the drive rods 180 degress opposed from left to right so they are always “pushing the wheel” like a bicycle crank?
The power’s not constant thru the revolution of the wheel, of course, but yes the piston does push and pull. The rods on the left side of the engine are 90 degrees out of phase with the rods on the right.
So if we’re looking at the left side of the engine, with the cylinders to our left–
when the crank is at 9:00 position, no tractive effort from the left side of the engine, but the right side of the engine is at 6:00 (or 12:00 on the PRR) so plenty on that side.
The pistons are double-acting. Steam is admitted to alternate sides of the piston, which is a flat disc, not a cylinder like an automobile engine. Also on a 2-cylinder steam locomotive one side is 90 degrees out of phase with the other,
so there is always a piston providing power.
And one thing to remember – the axles are solid so, although at one point the left side cylinder is pulling at its peak, and the right side one producing zero torque, both driving wheels (both sides) are pulling equally. So each axle gets four twists per revolution, on a two cylinder engine. Curiously, at full cutoff, the torque on the axles is very nearly constant (you work out the math if you like!).
The stresses (tending to rack the axles backwards and forwards in relation to the frame) are huge, but then, the frames are pretty heavy so it doesn’t matter – provided the bearings are good and the wedges adjusted properly.
I haven’t done the math… but about that. If I feel ambitious!.. At full cutoff (that is, valves wide open from beginning to end of the stroke – which never happens in reality), the torque is steam pressure x cylinder area (gives you pounds thrust on the main rods) x distance from axle centre to main rod bearing centre x (2 x abs(sin A) + 2 x abs(cos A) )where A is the angle or the wheel referenced to some logical point.
Not quite constant… varies by about plus or minus 20%.
Hmm… Well that helps me. All though I think I migth be a bit more confused than when I started… LOL
Actually, it makes sense.I figured there had to be something like that otherwise the wheel, er axle, would stop after the first rotation if the engine didnt move. Thats cool